# Fitch-Style Proof Help

I'm having some trouble solving a Fitch Proof, Here's how far I've gotten. Any Help is appreciated.

Thank You

• I don't really understand what you're doing, so I can't try to save your proof without understanding you. Your goal is $\forall x\left(\left(\text{Large}(x)\lor \text{Cube}(x)\right)\to \text{Dodec}(x)\right)$. The natural thing to do (for me at least) is to take a constant $a$ such that $\left(\text{Large}(a)\lor \text{Cube}(a)\right)$ and try to infer $\text{Dodec}(a)$. You haven't done this. So I ask you, what's your idea with the assumptions at steps 3. and 7.? – Git Gud Jun 11 '14 at 1:51
• I am trying to get Large(a) v Cube(a) → Dodec(a) is that possible or am I completely off track? – user3727903 Jun 11 '14 at 2:02
• It certainly is possible. At least you're being to asked to prove that. I just don't see how you intend to go there with your try. I will leave the layout of a proof in an answer,which hopefully you'll be able to formalize. Then I'll be off to bed. – Git Gud Jun 11 '14 at 2:05
• If this were a question in propositional calculus, then the use of the universal quantifier would come as quantifying over the domain of truth values. You can do this, however, your problem almost surely does not concern the propositional calculus with quantifiers, but rather predicate calculus. – Doug Spoonwood Jun 11 '14 at 2:13
• forums.philosophyforums.com/… – Doug Spoonwood Jun 11 '14 at 2:20

## 1 Answer

I don't get how you intend to go from what you've done to the goal. Let me suggest a different approach.

Your goal is $\forall x\left(\left(\text{Large}(x)\lor \text{Cube}(x)\right)\to \text{Dodec}(x)\right)$.

It's reasonable to take a constant $a$ such that $\text{Large}(a)\lor \text{Cube}(a)$.

Just because it is allowed, I suggest eliminating $\forall$ from the premises with $a$.

Now performing $\lor$-$\text{Elim}$ on $\text{Large}(a)\lor \text{Cube}(a)$ is one way to go.

If $\text{Large}(a)$ holds, then $\text{Dodec}(a)$ follows from the second premise (not exactly, but close enough).

If $\text{Cube}(a)$ holds, you can get a contradiction from the first premise and you can infer whatever you want, namely $\text{Dodec}(a)$.

Eliminating the disjunction and tying loose ends finalizes the proof.

Hover your mouse over the grey area below to see the suggested proof. 